CODEWITHSUNDEEP
c++stringuppercaselowercase
hilchev
hilchev

Reputation: 340

How does this array conversion work ( lowercase to uppercase using a string)?

I got the program to work as expected, but can anyone explain how it works?

#include <iostream>

using namespace std;

int main(void) {
    int exit;     
    string name;
    cin >> name;

    for (int i = 0; i < name.length(); i++) {
        // the line below is the one I don't understand 
        if ('a' <= name[i] && name[i] <= 'z') name[i] = char(((int)name[i]) - 32);
    }
    cout << name;
    cin >> exit;
    return 0;
 }

EDIT: Let me rephrase:

The thing I don't understand is how does the string-to-array deal work, as in: 'a'<= name[i]. What exactly does this compare and how?

EDIT2 Thanks for the quick responses guys, love you all. I figured it out.

Upvotes: 0

Views: 2484

Answers (5)

bames53
bames53

Reputation: 88215

if ('a' <= name[i] && name[i] <= 'z')

char objects are numeric values similar to ints. So 'a' <= name[i] is simply testing if the numeric value of 'a' is less than or equal to the character you're examining. Combined with name[i] <= 'z' and you're testing if the numeric value of name[i] is between the values of 'a' and 'z'. Now, it just so happens that the most common scheme for assigning numeric values to chars, named "The American Standard Code for Information Interchange" (ASCII), has the alphabet arranged in order; 'a' + 1 = 'b', 'b' + 1 = 'c', and so on. So figuring out if the character is between 'a' and 'z' tells you if it's a lower case letter.

name[i] = char(((int)name[i]) - 32);

Once you know that chars are just numeric values you might infer from the basic properties of arithmetic which we all learned in grade school that 'a' + ('A' - 'a') results in the value 'A'. Further, ASCII has the upper case alphabet arranged similarly to the lower case alphabet so that 'A' + 1 = 'B', etc. So taking anycharin the lower case alphabet and adding'A' - 'a'will result in the upper case version of that letter. In ASCII'A' - 'a'` happens to have the value -32. So take the numeric value for a lower case letter, subtract 32, and you have the value for the upper case letter.

For comparison here's a version of the code that doesn't depend on ASCII:

auto l = std::locale();

if (std::islower(name[i], l))
    name[i] = std::tolower(name[i], l);

Upvotes: 0

jxh
jxh

Reputation: 70472

I assume from the edit in your comment that you are wondering how the [] can apply to a string object. The operator [] is overloaded for string to return a reference to the character at the specified position offset of the represented string. There need not be any direct conversion of the string into an array. The code that implements the overload could well be walking a linked list. It depends on how string was implemented.

Upvotes: 2

David G
David G

Reputation: 96835

'a' <= name[i] && name[i] <= 'z'

This line is comparing the corresponding ASCII values of these two characters. 'a' in ASCII is 97 and 'z' is 122. If name[i] is one of the characters from 'a' to 'z' the expression returns true. This is commonly used to check if a variable is alphabetic.

Upvotes: 1

Yakk - Adam Nevraumont
Yakk - Adam Nevraumont

Reputation: 275740

This is the line:

 if('a'<=name[i] && name[i]<='z')name[i]=char(((int)name[i])-32);

broken down:

 if( 'a'<=name[i] ) {
   if( name[i]<='z' ) {
     // name_int is a temporary, which the above code implicitly creates,
     // but doesn't give a name to:
     int name_int = name[i];
     name_int = name_int - 32;
     name[i] = char(name_int);
   }
 }

and note that 32 happens to equal 'a'-'A' in the character encoding you are using.

(Technically name_int should be an int&& or somesuch, but no need to be that confusing.)

Upvotes: 3

Jesus Ramos
Jesus Ramos

Reputation: 23266

It assumes ASCII character format where to convert from lowercase to uppercase you subtract 32 from the original ASCII value. This is because the ASCII values for uppercase are smaller than those for lower case and it's a constant difference between A and a, B and b and so on.

For reference: http://www.asciitable.com/

Upvotes: 1

Related Questions